$I$ is incenter of $\triangle{ABC}$. The incircle touches $BC,CA,AB$ at $D,E,F$, respectively . Let $M,N,K=BI,CI,DI \cap EF$ respectively and $BN\cap CM=P,AK\cap BC=G$. Point $Q$ is intersection of the perpendicular line to $PG$ through $I$ and the perpendicular line to $PB$ through $P$. Prove that $BI$ bisect segment $PQ$.
Problem
Source: China South East Mathematical Olympiad 2016 Grade 10 Prob. 7
Tags: geometry, incenter
buzzychaoz
31.07.2016 10:22
First, it is well known that $M,N$ lie on $(BC)\implies P$ is the orthocenter of $\triangle BIC$, and that $AK$ bisects $BC$ (e.g. here), so $G$ is the midpoint of $BC$. Let $MN\cap BC=X$. Using $\triangle PBC$ as reference triangle, it is well known that $XI\perp PG$. Hence $-1=(X,D;B,C)=I(Q,P;BI\cap PQ,\infty)\implies BI$ bisects $PQ$.
rkm0959
31.07.2016 11:43
It's easy if you know your lemmas well! This is pretty much the above solution. Lemma 1. $G$ is the midpoint of $BC$. (Diameter of the incircle) Lemma 2. $M, N$ lies on $(BC)$. (Incircle chord perpendicularity) Lemma 2 gives us that $P$ is the orthocenter of $BIC$, so $I$ is the orthocenter of $BPC$, which results in $P, I, D$ colinear. Denote $T=EF \cap BC$. By Brokard on $(BC)$, we get that $PG \perp TI$. Perspectivity wrt $I$ gives the solution.
IAmTheHazard
27.10.2023 19:22
????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????? Use the Iran lemma and some other lemma whose name I forgot to get that $G$ is the midpoint of $\overline{BC}$, and $\angle BMC=\angle BNC=90^\circ$. Thus $I$ is the orthocenter of $\triangle PBC$ and the problem is actually asking the following. Actual problem wrote: Let $ABC$ be a triangle with orthocenter $H$. Let $M$ be the midpoint of $\overline{BC}$. Let the perpendicular to $\overline{AM}$ through $H$ intersect the perpendicular to $\overline{AB}$ through $A$ at $P$. Prove that $\overline{BH}$ bisects $\overline{AG}$. Let $D,E,F$ be the feet of the altitudes from $A,B,C$ respectively and let $T$ be the foot of the perpendicular from $H$ to $\overline{AM}$. Clearly $AEFHT$ is cyclic, and it is well-known that $BCHT$ is cyclic, so by radical center on $(AEFHT)$, $(BCEF)$, $(BCHT)$ it follows that $\overline{BC},\overline{EF},\overline{HT}$ concur at some point $X$. Then $$-1=(D,X;B,C)\stackrel{H}{=} (A,G;\overline{BH} \cap \overline{AG},P_\infty),$$and the conclusion follows. $\blacksquare$